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Where there's smoke...

Discussion in 'General Electronics Discussion' started by notTHEfoggiest, Jan 15, 2017.

  1. notTHEfoggiest

    notTHEfoggiest

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    Jan 14, 2017
    So I have been building a 200W LED ring light. In my attempt to improve upon the design in this video:



    Here are my main components used:

    20 x 27-30V 10W 300mA LEDs:

    http://www.ebay.co.uk/itm/161348456581?_trksid=p2057872.m2749.l2649&var=460852556717&ssPageName=STRK:MEBIDX:IT

    2 x Single core wired threaded together:

    http://www.hobbytronics.co.uk/breadboard-wire?keyword=single core breadboard wire

    1 x 10A 250W DC-DC step-up Voltage regulator

    Lipo alarm (not shown):

    http://www.ebay.co.uk/itm/191981053617?_trksid=p2057872.m2749.l2649&var=491530036174&ssPageName=STRK:MEBIDX:IT

    Vu LED:

    http://www.ebay.co.uk/itm/281226027217?_trksid=p2057872.m2749.l2649&ssPageName=STRK:MEBIDX:IT

    Wire used between the battery pack and DC-DC is thick 13A mains wire.

    [​IMG]
    I chopped and linished a knackered old frying pan to use as a heatsink/mount for the LEDs to make the build sturdier and able to run at higher power than the light in the video (looks too flimsy for my partner who breaks everything!). I mounted the LEDs with thermal paste in parallel using solid core wires braided together. I managed to burn out my first smaller 2-4A voltage regulator testing it with power coming from my converted ATX PSU using the 12V 14A rail (and no resistor - I can see how stupid this was now) but assumed I had connected the polarity incorrectly. Lesson 1 not yet learned.

    I ordered a beefier regulator:

    http://www.ebay.co.uk/itm/112218796285?_trksid=p2057872.m2749.l2649&ssPageName=STRK:MEBIDX:IT

    I then tested it again with 4 18650 batteries and both the batteries and the regulator got hot and the regulator gave off some smoke so I assumed it must be a short somewhere or the batteries are struggling. I found the wire running between the front to the back from the DC-DC were causing the short and re-did the solder this time with a thicker hole for it to go through. After I tested for continuity between both the pos/neg wire on the front and the pan it was showing open so all good.

    [​IMG]

    I then made a 6 cell battery pack to reduce the load on the cells and be more-like my final build (supply 1.2A or so each) and chopped some of the bolts on the back of the pan so I could mount the DC-DC on the back of the project. I tested for continuity between each pos and neg terminal on the DC-DC and the pan and it read as open too (resistance "1")
    [​IMG]
    [​IMG] [​IMG]

    You can see a voltage LED attached to the output. I then tested it this morning for a couple of seconds at a time at around 25V reading on the LED to find the DC-DC smoking a little again.

    Could the batteries be sending too much current/DC-DC be drawing too much? Already knackered another DC-DC? The breadboard wire x2 too skinny? It's rated 1.8A 1kV. I thought in a short run twined together, it would be fine?

    I'm waiting on a consignment of 30 sizes of 1/4W and 1/2W resistors to put one between the battery pack and the regulator however I have probably missed lots of crucial stuff. I would really appreciate some help in improving the build and finding the cause of the smoke. I have another DC-DC in the post (a few weeks off at least but was planned for another project). I'm very new to all of this so please excuse my rudimentary mistakes.

    Thanks
     
    Last edited: Jan 15, 2017
  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,172
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    Jan 21, 2010
    Are all of these LEDs in parallel? That's really not a good idea.

    Does the body converter you're using have a current limit?

    How long does this light have to stay on?

    How long was it on for before the boost converter let the smoke out?

    Did you measure the LED current? You're expecting 6A, right? And I'd this where you're using the wire rated for 2A or so?

    And what current are you expecting to draw from those batteries? About 12A? The larger regulator has a max input current of 10A.

    It sounds like you may be trying to do more than the boost regulator can do. Do you have specs on it? How hot was it getting? Do you know where the smoke was coming from?
     
  3. notTHEfoggiest

    notTHEfoggiest

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    0
    Jan 14, 2017
    The DC-DC in the video was rated 3-4A max however he probably used less powerful batteries. Yes the LEDs are in parallel, why is this a bad idea?

    I'm not sure, I think it only has V limit with a fine and rough pots 5 and 10k if I'm not mistaken?
    I plan for the light to be for 10-15 mins at a time, running around 1.5 hours on one battery charge however I would need to run lots of tests to see how hot it gets. Haven't got that far yet. The light heatsink never gets hot at all to the touch so far.

    I plan on running the LEDs at whatever setting they can be without getting hot to the touch as it will be mounted on the camera.

    Smoke comes out almost straight away.

    I didn't read the LED current as I was too busy unplugging the lead again lol.

    I thought cable was rated per metre? I thought slightly thinner wire would limit create more resistance and avoid too much current going to the lights. Am I totally mistaken?! The lights won't be running at full whack.

    The batteries are rated 2A each but I plan to draw like 1A from each battery which is within their rated limits by a large margin right? So the DC-DC is trying to draw everything?

    Here is the link to the DC-DC again:

    http://www.ebay.co.uk/itm/112218796285?_trksid=p2057872.m2749.l2649&ssPageName=STRK:MEBIDX:IT

    Here are the listed specs:


    Last time I tested it smoked with no heat. Do you think some current could have made it back to the DC-DC from the screws I have it sitting on or mounted to? They are in contact with the pan but I couldn't see anywhere it's connected to the live circuit. The smoke seemed to come from the coil on the DC-DC but I could be mistaken.

    Thanks for all the help
     
    Last edited: Jan 15, 2017
  4. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,172
    2,689
    Jan 21, 2010
    In our resource section we have a resource about driving LEDs. Please read it. In your case I'd recommend you read all of it.

    Without a current limit there's nothing to prevent you from delivering too much current to the LEDs. If you've read and understood the first section of the previously mentioned resource you should understand at least part of the reason why.

    By my rough calculations this requires each of your cells to be about 20,000mAH (20Ah). I'd that what they are?

    Given some of your other answers I hope this is your wires smoking. That is the least worst option.

    I would have had one meter reading the output voltage and another reading the current. I would have had the output voltage set to the minimum and carefully turned it up while monitoring the output voltage and current.

    Pretty much, yes.

    Wire need to be rated to carry the current required or it will get hot. Too hot will mean the insulation melts or burns. Artist from the possible smell, mess, and risk of fire, the is the additional problem of the exposed wire then shorting out.

    If you need resistance, use resistors. Note that you will need to ensure that the resistors are rated for the power they will be called on to dissipate.

    Lots more misunderstanding he too.

    You have 20 LEDs in parallel. Each draws 0.3A at full power. That's 6A at maybe 33V.

    You mentioned you were using a 12V supply to drive the boost converter. Assuming about 80℅ efficiency that's about 250W into the boost converter for 200W out. At 12V that's a little in excess of 20A

    If you are using 6 lipo cells, the input voltage is probably closer to 20V under load, so the current will be lower, probably around 12A.

    The batteries are in series, so the full current is required from each cell.

    If you expect 1A and you're drawing 6A, that may explain why your batteries are getting warm!

    The dc-dc converter draws what is required. I think you can see your calculations may have been a little off.

    It's possible that you've sorted something when mounting the regulator but then if not expect the LEDs to illuminate. However, given that your using wire that is way too thin, its resistance may be sufficient to limit the current (causing overheating and smoke)

    I presume you checked everything carefully with a multimeter before you Peter it on to make sure the were no shorts?

    Honestly, I would be measuring both input and output current and voltage on that boost converter when I powrred it up. However I imagine you don't have 2 (let alone 4) multimeters.

    In your case, I would recommend that you first recalculate the required voltages and currents to get ballpark figures.

    I would recommend you get a DC-DC converter with a current limit. If it's coming from China if recommend keeping it at or below half of its rated current and power. If this requires you to have multiple converters, that's not necessarily a bad thing.

    If you're running these LEDs in parallel, if recommend that each has a resistor in series with it that drops maybe 1V at the rated current. If probably use a 3.3 ohm resistor. You could get away with a 1/2W resistor here. This resistor will not limit the current, but it should help balance the current to each LED.

    You're also going to need to review the required capacity of your battery pack.
     
  5. notTHEfoggiest

    notTHEfoggiest

    17
    0
    Jan 14, 2017
    I've started reading the thread (very well put into layman's/idiot terms so far thank god!) and will read it thoroughly and digest/ask questions about it before doing anything else practical on the project.

    I was hoping to have the LED's running well under their max rated specs, I suppose a bit more optimistic than yourself! By your calculations I could get 50 mins as mine came out on average 1800 mAh or thereabouts which isn't too shabby at all. If using the light inside, the main worry is going to be making sure I sort out my understanding and control of the amps as outlined in your helpful FAQ section.

    I wasn't sure which way would increase the resistance on the 10 and 5k pots, when I last used it they were at 25v output so I hoped I could reduce it more as soon as I plugged it in (monitoring the V on the Vu LED) but alas this didn't work out. I actually do have another LED Vu and two multimeters but only two pairs of hands. I think a lack of tools wasn't the main issue here if we're honest...

    I think we have had some crossed wires here (see what I did there?). I said I tested the little supply once on the ATX PSU once to it's destruction. However, since then, I have only used the 18650 batteries.

    So re the batteries: Would it be preferable to use 8 batteries and put them in pairs/groups of parallel? I read the pitfalls of this method (i.e all failing when one does) and wasn't planning to do this but I misunderstood how the series circuit would work re: amps. I didn't realise they would need to give out the full amount of amps each which is quite a lot and not what they were designed for long periods I'm guessing.

    Here are the values after I tested them on the accucell 6 although I only did one cycle and I didn't connect it to the computer to see the voltage curve whilst discharging:

    Battery 1 : 2001 mAh
    Battery 2 : 1990 mAh
    Battery 3 : 1978 mAh
    Battery 4 : 1669 mAh
    Battery 5 : 1644 mAh
    Battery 6 : 1636 mAh
    Battery 7 : 1600 mAh
    Battery 8 : 1600 mAh

    The battery holders I bought are so bad and with puny wires so I was only using them to test and was planning to solder thick wires to them to avoid using the puny leads and connectors that seem prevalent in most cases I have used/seen so far. I am aware I have to be careful when doing this and prime the batteries to take the solder so I planned to again use the knackered old laptop batteries as expendable test subjects.


    They did illuminate... I thought the backplate wouldn't need an insulator as it's designed to go on a heatsink? It should already be insulated from the dc-dc right?


    I checked as best I could. I checked resistance between each wire on the LED's and the pan and on the connections on the dc-dc. I think I will need to read up further on multimeter usage, I have seen a fair few video tutorials but not nearly enough it would seem...


    I think the largest deciding factor would be the battery setup and the battery count before I can sort this part out?


    Ok, my reading of your LED section is necessary here too. Resistors of 30 types in 1/4 and 1/2W are on their way luckily.

    Thanks for all the food for thought, I am constantly trying to order stuff asap due to the length of time it takes to get here. I have yet to see people who sell stuff "in the UK" who are not simply bulk-buying and up-selling the stuff made in China. I think I will need to find out after my reading whether my dc-dc has snuffed it and then see if what the solution is to getting a constant current. I will say now that I'm not planning on buying more expensive and elaborate stuff as this is a project after all. I hear your advice so will try and find a compromise between the perfect solution and my wallet. :D
     
    Last edited: Jan 16, 2017
  6. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    In series connection, which is best for a number of reasons, you need to consider the lowest capacity. So in this case you have 8 cells with a usable capacity of 1600mAh.

    You can't use all the cells in series because the voltage is too high. If you arrange them as 4 pairs of 2 cells in series, you have a battery of nominally 14.4V and 3200mAh capacity.

    If look at a 1 hour discharge rate, (3.2A) which will probably give you a realistic 50 minute life, your budget is about 45W.

    Allowing for 80% inverter efficiency, you have 37W to drive the LEDs. If this is all delivered to the LEDs, that's about 1.8W per LED.

    You'll be operating them at about 20% brightness.

    Try one of them at 60mA and see if that seems bright enough for you.
     
  7. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    Set the voltage before you connect the load.

    Use a multimeter to measure the voltage. Remember though, it's current that really matters and will vary even if the voltage starts constant.
     
  8. notTHEfoggiest

    notTHEfoggiest

    17
    0
    Jan 14, 2017
    How does that maths work? Wouldn't you do 4 x the capacity of 1600 just like you have the voltage? I make that out as 6400?

    I have read the article on LED's and for the most part I get the main ideas however when it comes to the circuit diagrams I get easily lost. The video author has got a V rather than A regulator. I see and understand the overheating aspect. Ok...

    In my mind it seems I need to order the following:

    http://www.ebay.co.uk/itm/142195893922?_trksid=p2060353.m1438.l2649&ssPageName=STRK:MEBIDX:IT

    and potentially 8 more 18650 cells to give me more running time. It would seem that it won't last at all long outside. I will set it up to run on mains as this light is not really designed for outside all that much anyways. I can always either use the battery pack for however long it lasts (not very long) or use it inside with the PSU.

    Where I get a bit hazy is the voltage, in the info it says the perfect solution is unlimited resistance and voltage however could not limiting the voltage potentially damage the LEDs or is it only the current that can do this? Using the above current regulator would give far higher tolerance and lower running temperature of the regulator than another running slightly closer to capacity right? Or do I still need something bigger?

    Or is there a cheaper, better solution? I surmised that using fixed resistors don't solve the thermal problem fully and waste more energy right?

    It would seem that testing it is a dangerous thing to do before using resistors or a current regulator?

    Do you think I could get away with soldering on another 3 wires twined together on top of the existing ones on the LEDs? I hear you re: the possibility for shorting however I had great difficulty desoldering one of the LEDs to drill thicker holes. Also, using an insulated cable will end up looking awful after I have cut into it and melted the insulation soldering the wire. Maybe I could add insulation after? That way I avoid the ball ache of desoldering, potentially damaging the LEDs and it should look better too?
     
  9. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,172
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    Jan 21, 2010
    You can add the mAh ratings when cells are in parallel, you add the voltage when they are in series.

    Each battery stores a certain amount of energy. This is approximately the voltage time the Ah rating. In your case it's about 1.6 * 3.6 = 5.76Wh. (let's call it 5.75Wh).

    Any valid arrangement the cells into a battery will have a total Wh rating equal to the number of cells times their individual energy capacity. For 8 cells that's 46Wh.

    If you divide 46Wh by the nominal battery voltage (in this case 14.4V) you get the Ah rating. In this case that is 3.2Ah, or 3200mAh.

    This piece of confusion is probably THE most common misunderstanding about batteries.
     
  10. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    That looks to be a good solution. It is reasonably overrated as long as you keep the input voltage high enough.

    I should probably remove that statement. It is theoretically correct, but not easily interpreted.

    If the current is correct, the voltage will not rise too high. However, if you are able to set both the voltage and the current IN THIS CASE you set the current to the correct value and the voltage to a value a little higher than is required (if you need 32V, set it for 35V or so). If the LEDs are removed, this sets an upper limit on the open circuit voltage.

    That is exactly correct. However, in your case where you have several LEDs in parallel, a small amount of resistance in series with each of them will help balance the current.

    You need to be very careful. You have to constantly monitor the LED current because it will change very quickly with a small change in voltage. A slight tweak can easily move you from LEDs and regulator comfortably within their ratings to a current which overstresses the LEDs and the regulator. Because things will change as the LEDs heat up, testing at higher powers will be a dynamic exercise requiring you to reduce the voltage slightly as the LEDs heat up to maintain a safe current.

    Yes. A short section of thinner wire is OK. Another solution is to power the string from several positions around the loop (and the same for the ground connection. Running extra wires around the loop may be the easiest. If i is a complete loop (i.e. both ends are connected) then you need to make the wire thick enough to carry 1/2 the full current. If you feed power in 2 places on a closed loop, the loop need only be sized to carry 1/4 the current.

    The next thing you need to be concerned about is over-discharging your batteries. You will need to monitor them and turn off the light when any cell drops below about 2.8V (preferably a little higher)
     
  11. notTHEfoggiest

    notTHEfoggiest

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    Jan 14, 2017
    Oh balls.
     
  12. notTHEfoggiest

    notTHEfoggiest

    17
    0
    Jan 14, 2017
    Well the rails on my PSU are:

    +3.3V 21A
    +5v 20A
    +12V1 14A
    +12VV2 14A
    +5Vsb 2.5A
    -12V 0.5A

    with max output for both 12V rated at 288W

    Could I combine all the yellow 12V1 and 2 cables for one large 12V output?
     
  13. notTHEfoggiest

    notTHEfoggiest

    17
    0
    Jan 14, 2017
    If the DC-DC I get only has setting for current (looks likely) is this going to be sufficient?

    What are you thinking to define "small"? So a resistor between every LED on both pos and neg wires? If this were you would you desolder the bastards or chop a section out in the wires around the front of the LED?


    Remember I don't want it to be full brightness, it's blindingly bright so hopefully if I can have it drawing less amps, let's use an arbitrary figure like 4A then shouldn't that take out a lot of the overheating problems? It needs to be usable without any thought other than "on" and "off" as I'm building it for my partner to use.

    Come again? Having current coming from two sides reduces the thickness of the wire needed? What the? But the wires are still carrying the same current? What kind of sorcery is this?! So I could simply add another couple of wires to the output of the DC-DC and run them to the opposite side of the light and I won't have to add extra wire? Why did't you say this in the first place?! lol

    Is this because the LED's effectively use the current before it gets the the middle of the wire?
     
  14. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    I'm not sure the computer power supplies like that. Generally speaking they shut down if anything is badly wrong so you might be OK doing this.

    Again, no guarantee, but it certainly looks a lot better. I'd be pretty confident.

    I'll try to answer the rest later...
     
  15. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    I think I suggested a 10ohm 1/2W resistor.

    You need one for each LED, but only in series with one lead (either one, it makes no difference).

    In that case you'll also have to remove one of the wires going around the LEDs (sorry).

    I'd still design it so it is capable of running at full power. Turning it down will make it run cooler and be less likely to fail. It's best to have lots of margin, especially with stuff from China where you can be almost sure that it has no margin when run at full power.

    Remember that each LED takes some current, if the parallel LEDs were in a single line fed from one end, the wires leading up to the first LED carry 100% of the current. 5% gets shunted through the first LED, leaving 95% to go along the wires to the second LED which shunts another 5%, leaving 90% going to the third... And eventually at the end, 5% remains for the 20th LED.

    If the LEDs are arranged in a circle, the full current goes along the feed wires to the ring of LEDs. Assuming it is connected between the LEDs 1 and 20, 50%of the current goes either way, and at the opposite side of the loop (theoretically) zero current flows between LEDs 10 and 11.

    If you feed the circle at 2 points, the current is split 4 ways.

    Note to self... Read the whole post :) Yep, you got it right.
     
  16. notTHEfoggiest

    notTHEfoggiest

    17
    0
    Jan 14, 2017
    [​IMG]

    Like this around the whole length? I suppose I will get a good opportunity to practice my soldering skills!

    I like the way you think, design for the worst case scenario and hope for the best.

    You know what, for the first time this is what I would have assumed (being logical) but told myself that I must be wrong as so much of the time what I think is logical doesn't work in electronics! I suppose if I solder on the extra cable anyways whilst I'm there I'm adding in extra headroom (it's the de-soldering that's a ball ache).

    I managed to finish my SA tax return ahead of schedule so assuming I can get my admin for next week in order this afternoon I'm going to give this another crack tonight or tomorrow.

    Have a look what came through today and yesterday, my mailman is convinced that I'm building a bomb with the amount of small parcels I get through at the moment!
    [​IMG]
     
  17. BobK

    BobK

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    Jan 5, 2010
    Nope, the resistors go between the ring carrying power and the LED devices. One for each LED on either of the two contacts.

    Bob
     
  18. notTHEfoggiest

    notTHEfoggiest

    17
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    Jan 14, 2017
    Hi Bob, you mean like my picture (between each LED) but on the inner Negative wire?
     
  19. BobK

    BobK

    7,592
    1,636
    Jan 5, 2010
    No, between the ring and the LED. Your picture shows the resistors breaking up the ring.

    Bob
     
  20. notTHEfoggiest

    notTHEfoggiest

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    Jan 14, 2017
    Bob you're going to need to draw a picture. There are many rings on the light and so I am no closer to understanding where the resistor needs to go! haha
     
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